What happens when an object is dropped from a very tall tower? The critics against Copernicus and Galileo argued that, if the earth moved, then, a heavy body that was dropped from a very high tower should fall to the west of the foot of the tower. Galileo argued, however, that this body would fall slightly to the east. My first guess is that the body would fall directly under the point in which it was dropped. Who is right?  
 A: When the ball is being held "stationary" at the top of the tower, then it is moving with the same angular velocity as the Earth. However, this does not mean that its angular velocity will remain constant during the drop. Indeed, it can't: because the object is moving at a nonzero radius from the axis of the Earth's rotation, it has angular momentum, and that angular momentum must be conserved. To account for the decrease in radius from the axis of rotation, the angular velocity of the object's rotation about the Earth's axis must increase. Therefore, the rotation of the object "speeds up" as it falls. Therefore, it rotates "faster" than the Earth under it, and so it falls slightly to the east.
In summary, the object lands to the east of the point over which it is dropped, as a consequence of the conservation of angular momentum.
A: 
Who is right?

Galileo.
This sounds a little like a homework problem. Is it? If so it should be labelled as such.
The sun rises in the east and sets in the west. Thus the earth is rotating "to the east". The earth ostensibly has a constant rotational velocity $\omega$ of about one revolution per 24 hours. That means a point at the base of the tower has a linear velocity "to the east" of 
$$
v_{tower}=R_e\omega\;,
$$
where $R_e$ is the radius of the earth.
And a point at the top of the tower (of height $H$) has a velocity "to the east" of
$$
v_{body}=(R_e+H)\omega\;,
$$
which is also the initial velocity of the dropped body.
You should be able to work out the rest on your own.
